How do i find a closed form expression for $\sum_{k=0}^n \frac{(x-1)^k}{k+1}$?

Question

How do I Find a closed form expression for :

$$\sum_{k=0}^n \frac{(x-1)^k}{k+1}$$

Note :I have no idea how to do that, I am bad at evaluating series when we cannot use some standard series to do it.

Thank you for any help

Answer 1

We begin with the finite geometric series $f_n(z) = \frac{z^{n+1} - 1}{z-1} = \sum_{k=0}^n z^k. $ Integration term-by-term with respect to $z$ then gives $ \int f_n(z) \, dz = \sum_{k=0}^n \frac{z^{k+1}}{k+1} = z \sum_{k=0}^n \frac{z^k}{k+1}.$ Therefore, $S = \sum_{k=0}^n \frac{(x-1)^k}{k+1} = \frac{1}{x-1} \int_{z=0}^{x-1} f_n(z) \, dz .$ This integral does not have an elementary closed form antiderivative. It is, however, expressible as an incomplete beta function: $S = -\frac{\log(2-x) + {\rm B}(x-1;n+2,0)}{x-1}.$ That's the best you can really hope to do, but it's really no different than writing $S$ in terms of the integral of $f_n$. If this question is part of a larger question and you are certain it is posed correctly, chances are that you are approaching the solution incorrectly

Answer 2

$$\sum \frac{(x-1)^k}{k+1}=\\\frac{1}{x-1}\sum \frac{(x-1)^{k+1}}{k+1}\\$$ Suppose \begin{align} S &= 1+(x-1)+(x-1)^2+(x-1)^3 +...+(x-1)^k \\ &=(x-1)^0+(x-1)+(x-1)^2+(x-1)^3 +...+(x-1)^k \end{align} now integral of s: \begin{align} \int Sdx &= \int ((x-1)^0+(x-1)+(x-1)^2+(x-1)^3 +...+(x-1)^k) \\ &= (x-1)+\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}\frac{(x-1)^4}{4}+...+\frac{(x-1)^{k+1}}{k+1} \end{align} note that S is sum of geometric progression

Answer 3

Let $f(x) = \sum_{k=0}^n \frac{(x-1)^{k}}{k+1}\text{ and } 1 < x < 2$, $$F(x) + C= \int f(x)\ dx = \int \sum_{k=0}^n \frac{(x-1)^{k}}{k+1}\ dx$$ $$= \sum_{k=0}^n \int\frac{(x-1)^{k}}{k+1}\ dx = \sum_{k=0}^n (x-1)^{k+1}$$$$ =(x-1)\cdot\sum_{k=0}^{n}(x-1)^k = (x-1)\cdot\frac{(x-1)^{n+1} - 1}{(x-1) - 1}$$ $$f(x) = \frac{d}{dx}F(x) = \frac{\big[(n+1)(x-1) - (n+2)\big](x-1)^{n+1} - 1}{(x-2)^2}$$